Bài 1:

Vì $a\geq 1$ nên:

\(a+\sqrt{a^2-2a+5}+\sqrt{a-1}=a+\sqrt{(a-1)^2+4}+\sqrt{a-1}\)

\(\geq 1+\sqrt{4}+0=3\)

Ta có đpcm

Dấu "=" xảy ra khi $a=1$

Bài 2:
ĐKXĐ: x\geq -3$

Xét hàm:

\(f(x)=x(x^2-3x+3)+\sqrt{x+3}-3\)

\(f'(x)=3x^2-6x+3+\frac{1}{2\sqrt{x+3}}=3(x-1)^2+\frac{1}{2\sqrt{x+3}}>0, \forall x\geq -3\)

Do đó $f(x)$ đồng biến trên TXĐ

\(\Rightarrow f(x)=0\) có nghiệm duy nhất

Dễ thấy pt có nghiệm $x=1$ nên đây chính là nghiệm duy nhất.

\(a,PT\Leftrightarrow x^2-3x+2+x^2-x\sqrt{3x-2}=0\left(x\ge\dfrac{2}{3}\right)\\ \Leftrightarrow\left(x^2-3x+2\right)+\dfrac{x\left(x^2-3x+2\right)}{x+\sqrt{3x-2}}=0\\ \Leftrightarrow\left(x^2-3x+2\right)\left(1+\dfrac{x}{x+\sqrt{3x-2}}\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)\left(1+\dfrac{x}{x+\sqrt{3x-2}}\right)=0\)

Vì \(x\ge\dfrac{2}{3}>0\Leftrightarrow1+\dfrac{x}{x+\sqrt{3x-2}}>0\)

Do đó \(x\in\left\{1;2\right\}\)

\(b,ĐK:0\le x\le4\\ PT\Leftrightarrow x+2\sqrt{x}+1=6\sqrt{x}-3-\sqrt{4-x}\\ \Leftrightarrow x-4\sqrt{x}+4=-\sqrt{4-x}\\ \Leftrightarrow\left(\sqrt{x}-2\right)^2=-\sqrt{4-x}\)

Vì \(VT\ge0\ge VP\Leftrightarrow VT=VP=0\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{4-x}=0\end{matrix}\right.\Leftrightarrow x=4\left(tm\right)\)

Vậy PT có nghiệm \(x=4\)

a: =>3x+3=4x-4

=>-x=-7

hay x=7(nhận)

b: (x-1)(x-3)=0

=>x-1=0 hoặc x-3=0

=>x=1 hoặc x=3

c: 2(x-1)+x=0

=>2x-2+x=0

=>3x-2=0

hay x=2/3

a, ĐKXĐ : x ≠ 1 ; x ≠ -1

\(\Rightarrow3\left(x+1\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x+3=4x-4\)

\(\Leftrightarrow-x=-7\)

\(\Leftrightarrow x=7\left(N\right)\)

b,

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

c,

\(\Leftrightarrow2x-2+x=0\)

\(\Leftrightarrow3x=2\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

\(4\left(x+1\right)^2=\sqrt{2\left(x^4+x^2+1\right)}\)

\(\Leftrightarrow16\left(x+1\right)^4=2\left(x^4+x^2+1\right)\)

\(\Leftrightarrow\left(x^2+3x+1\right)\left(7x^2+11x+7\right)=0\)

\(\sqrt{\frac{x+56}{16}+\sqrt{x-8}}=\frac{x}{8}\)

\(\Leftrightarrow2\sqrt{x+56+16\sqrt{x-8}}=x\)

\(\Leftrightarrow2\sqrt{\left(\sqrt{x-8}+8\right)^2}=x\)

\(\Leftrightarrow2\sqrt{x-8}+16=x\)

\(\Leftrightarrow x=24\)

ĐK \(x\ge-4\)

\(BPT\Leftrightarrow\hept{\begin{cases}2x-3\ge0\\x\ge-4\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge\frac{3}{2}\\x\ge-4\end{cases}}\)

\(\Rightarrow x\ge\frac{3}{2}\)

ĐK: \(x+4\ge0\) <=> \(x\ge-4\)

Bpt <=> \(\orbr{\begin{cases}x+4=0\\2x-3=0\end{cases}}\) hoặc \(2x-3>0\) <=> \(\orbr{\begin{cases}x=-4\\x=\frac{3}{2}\end{cases}}\)hoặc \(x>\frac{3}{2}\)

<=> \(\orbr{\begin{cases}x=-4\\x\ge\frac{3}{2}\end{cases}}\)Thỏa mãn đk.

Vậy 

\(\orbr{\begin{cases}x=-4\\x\ge\frac{3}{2}\end{cases}}\)

a: 2/(x-2)=3/(x+2)

=>3x-6=2x+4

=>x=10

b: (x-2)(x+5)=0

=>x-2=0 hoặc x+5=0

=>x=2 hoặc x=-5

c: 2(x+2)-x=4

=>2x+4-x=4

=>x=0

\(a,\dfrac{2}{x-2}=\dfrac{3}{x+2}\)

\(\Leftrightarrow\dfrac{2}{x-2}-\dfrac{3}{x+2}=0\)

\(\Leftrightarrow\dfrac{2\left(x+2\right)-3\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow2x+4-3x+6=0\)

\(\Leftrightarrow-x+10=0\)

\(\Leftrightarrow-x=-10\)

\(\Leftrightarrow x=10\)

\(b,\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

\(c,2\left(x+2\right)-x=4\)

\(\Leftrightarrow2x+4-x-4=0\)

\(\Leftrightarrow x=0\)

Bởi vì \(\sqrt{2x+1}\ge0\)mà \(x>\sqrt{2x+1}\)nên phải có điều kiện \(x>0\)